Epipolar Geometry
Computer Vision CMP-6035B
Dr. David Greenwood
Spring 2022
Contents
- Motivation
- Epipolar Geometry
- Epipolar Elements
- Point Correspondences
Motivation
Given \(x'\) in the first image, find the corresponding point \(x''\) in the second image.
- Coplanarity constraint
- Intersection of two corresponding rays
- The rays lie in a 3D plane
Epipolar Geometry
- describe geometric relations in image pairs
- efficient search and prediction of corresponding points
- search space reduced from 2D to 1D
Epipolar Geometry
Epipolar Geometry
Epipolar elements:
- epipolar axis \(\mathcal{B} = (O' O'')\)
- epipolar plane \(\mathcal{E} = (O' O'' X)\)
- epipoles \(e' = (O'')', e'' = (O')''\)
- epipolar lines \(\mathcal{L}'(X) = (O'' X)', \mathcal{L}''(X) = (O' X)''\)
We can also write the epipoles as:
\[ e' = (O' O'') \cap \mathcal{E}', \quad e'' = (O' O'') \cap \mathcal{E}'' \]
And the epipolar lines as:
\[ \mathcal{L}'(X) = \mathcal{E} \cap \mathcal{E}', \quad \mathcal{L}''(X) = \mathcal{E} \cap \mathcal{E}'' \]
In the Epipolar Plane
Assuming a distortion free lens:
- The projection centres \(O'\) and \(O''\).
- The observed point \(X\).
- The epipolar lines, \(\mathcal{L}'(X)\) and \(\mathcal{L}''(X)\).
- The epipoles, \(e'\) and \(e''\).
- The image points \(x'\) and \(x''\).
In the Epipolar Plane
- The projection centres \(O'\) and \(O''\).
- The observed point \(X\).
- The epipolar lines, \(\mathcal{L}'(X)\) and \(\mathcal{L}''(X)\).
- The epipoles, \(e'\) and \(e''\).
- The image points \(x'\) and \(x''\).
All lie in the epipolar plane \(\mathcal{E}\).
Predicting Point Correspondence
Task: Predict the location of \(x''\) given \(x'\).
- For the epipolar plane \(\mathcal{E} = (O'O''X)\)
- The intersection of \(\mathcal{E}\) and the second image plane \(\mathcal{E}''\) yields the epipolar line \(\mathcal{L}''(X)\)
- The corresponding point \(x''\) lies on that epipolar line \(\mathcal{L}''(X)\).
- Search space is reduced from 2D to 1D.
Computing the Elements of Epipolar Geometry
Computing the Elements of Epipolar Geometry
- We described the important elements geometrically.
- We will compute them using the projection matrices and the fundamental matrix.
Epipolar Axis
The direction of the epipolar axis can be derived directly from the projection centres.
\[ b = X_{O'} - X_{O''} \]
The vector \(b\) is homogeneous, we know only the direction, not the length.
Epipolar Lines
Image points lie on the epipolar lines.
\[ x' \in \mathcal{L}' , \quad x'' \in \mathcal{L}'' \]
Epipolar Lines
For \(x'\):
\[ x^{'T} \mathcal{L}' = 0 \]
Epipolar Lines
For \(x'\):
\[ x^{'T} \mathcal{L}' = 0 \]
We can exploit the coplanarity constraint for \(x'\) and \(x''\):
\[ x^{'T} \underbrace{Fx''}_{\mathcal{L}'} = 0 \]
Epipolar Lines
We can exploit the coplanarity constraint for \(x'\) and \(x''\):
\[ x^{'T} \underbrace{Fx''}_{\mathcal{L}'} = 0 \]
\[ \mathcal{L}' = Fx'' \]
Epipolar Lines
The same for \(x''\):
\[ \mathcal{L}^{''T} x'' = 0 \]
We can exploit the same constraint \(x^{'T}Fx''= 0\) and obtain:
\[ \mathcal{L}^{''T} = x^{'T}F \]
Epipolar Lines
We can exploit the same constraint \(x^{'T}Fx''= 0\) and obtain:
\[ \mathcal{L}^{''T} = x^{'T}F \]
\[ \mathcal{L}^{''} = F^{T}x^{'} \]
Epipolar Lines
Image points lie on the epipolar lines, \(x' \in \mathcal{L}'\) and \(x'' \in \mathcal{L}''\).
- we can exploit the coplanarity constraint \(x^{'T}Fx''= 0\).
- which is valid if:
\[ \mathcal{L}' = Fx'' \quad \mathcal{L}'' = F^{T}x' \]
Epipoles
The epipoles are the projection of the camera origin onto the other image.
- Both can be computed using the projection matrices.
Epipoles
The epipoles are the projection of the camera origin onto the other image.
- Both can be computed using the projection matrices.
\[ e' = P'X_{O''} \quad e'' = P''X_{O'} \]
Epipoles
The epipole is the intersection of all the epipolar lines in an image.
\[ \forall \mathcal{L}' : e^{'T} \mathcal{L}' = 0 \quad \forall \mathcal{L}'' : e^{''T} \mathcal{L}'' = 0 \]
Summary
- We assumed an uncalibrated camera.
- We discussed Epipolar geometry and epipolar elements.
- Epipolar geometry reduces the correspondence search from 2D to 1D.
Reading:
- Forsyth, Ponce; Computer Vision: A modern approach.
- Hartley, Zisserman; Multiple View Geometry in Computer Vision.
Data: